On the Drawing of Figures of Crystals. 39 
If the eye be elevated above the horizontal plane, the lines P Y’, 
NU, HZ will be projected below GG. The lengths of these 
projections are in direct ratio to the lines projected. 
To obtain the values of PY’, NU’, HZ, we observe that 
PY’=cosd; NU’=cos (60°—5); HZ=cos (60°+ 4). 
Whence, since cos=4/(? —sin?), 
cosd=/(1—sin?d) = 5734; 
cos (60° —d)=/ (1 —4 sin? 9)= 4735; 
cos (60°+5)=/(1—9 sin? ))= VY z;- 
From these equations, the following relations result: 
PY: :AZ::5:4:1, 
which, therefore, is also the ratio of the projections of these lines 
below GG consequent on an elevation of the eye at any angle «. 
1 
If the second projection of the semiaxis I (y’P), = ~ part of PM 
(the first projection), 
cot e=scotd=5s V1. 
For if P Y’ be made radius in the two triangles P Y’y/ and P YM, 
we shall have Py’=tan «, and PM=tand. But Py’: PM:13:s; 
consequently tane ; tand::1: s; 
tan 0 
tan e= eat 
-. cot e=scot 9, 
In general it is most convenient to assume 2 as the value of s; 
then e=9° 50’. If s= 4/3; #=11° 18°5’. 
14. Projection of the axes——The above demonstration affords 
a method of projecting the tetraxonal axes, which is similar to the 
method in the monometric system. We may assume r=3, s=2. 
1. Draw the lines AA, HH Fig. 7. 
at right angles with, and bisect- Z 
ing each other. Let HM=6, i S$) .g24 gee 4 
or HH=2b. Divide HH into 
six parts by vertical lines. These % 
lines including the left and right ¥ 
hand verticals may be numbered aa ae a 
from one to six as in the figure. a 
In the first vertical, below H, * 
‘ 
lay off HS=~ 6, and from S , 
~~ 
x fe 
